If I have a metal ring and I heat it up, I will expect thermal expansion.

From my point of view of expansion, the metal will expand to free space. In that case, the inner radius of the ring should be smaller and the outer diameter of the ring should be bigger.

I understand that is not the case, but could anyone explain me why?


5 Answers 5


Let's suppose that the inner radius is actually becoming smaller.

This would mean that the molecules on the inner radius would have to be "packed" closer.

(also applies to any molecules that reside on the radius smaller than $\dfrac{R_1 + R_2}{2}$)

But then it would contradict with the fact the metal is expanding. Hence the inner radius will have to remain at least of its initial size.

Below is the pic of the expansion you suppose to happen (it won't happen this way, for the reasons described above).

  • $\begingroup$ This doesn't fully address the question. You've shown that the inner radius has to be least of the same size than initially. You did not explain why it will actually increase its size, which is mentioned in the question. $\endgroup$ Apr 7, 2018 at 12:24
  • $\begingroup$ @no_choice99 I did address the fact OP couldn't actually understand (the question title needs editing; just see the body of the question). We had a brief discussion in comments here, so OP is satisfied with my answer (they wouldn't accept otherwise, of course.) $\endgroup$
    – nicael
    Apr 7, 2018 at 12:30
  • $\begingroup$ Of course he is entirely satisfied, but not necessarily someone who has the same question than him. In the body text of the OP's post, one reads "(...) but why the inner radius will become larger?" This question is left unanswered by your answer. $\endgroup$ Apr 7, 2018 at 13:16
  • $\begingroup$ @nicael : you might be interested in this. $\endgroup$ Apr 8, 2018 at 13:19

Another way to answer this question is based on the assumption that we are dealing with a linear expansion.

This means that the distance between any two points in a heated body will increase by the same percentage.

So if our two points are the end points of the internal diameter of a heated ring, we'll have to conclude that the internal diameter will expand... by the same percentage as the external diameter or the distance between any other two points of the ring.

  • $\begingroup$ That's a better answer than the accepted one and currently top voted. $\endgroup$ Apr 7, 2018 at 12:25
  • $\begingroup$ @no_choice The question "why does it expand" is not an original problem, since that's what OP was told already. But OP didn't understand one thing, why wouldn't it expand in both directions (i.e., the outer radius growing and the inner radius shrinking). I've provided an intuitive explanation of why wouldn't the inner radius shrink, since it is something that OP had a problem understanding. $\endgroup$
    – nicael
    Apr 7, 2018 at 12:40
  • $\begingroup$ @V.F. answers the OP's question "(...) but why the inner radius will become larger?". As it is currently written, your answer (@nicael), doesn't address this question. $\endgroup$ Apr 7, 2018 at 13:18
  • $\begingroup$ I don't understand one thing.. Shouldn't the ring expand a really instead of linearly. I think only area is expanding $\endgroup$
    – Scáthach
    Sep 13, 2018 at 5:27

Yet another way to think about this situation, in addition to the good answers so far, is: imagine you are uniformly heating a solid disk that has many concentric circles inscribed upon it. As you heat the disk, all the circles get bigger. Moreover, the disk does not buckle or fold anywhere from an internal stress. It smoothly and uniformly gets bigger everywhere.

Now suppose you cooled the disk and removed the material inside one of the circles. If you heated up the now-ring again, the same thing would happen as before; all the remaining circles would get bigger, including the one that is now the interior edge of the ring. The ring doesn't "know" that there's no center in the disk anymore; it gets bigger just like it did before.


what you describe is a standard job interview question traditionally inflicted upon materials scientists. here is the "standard" answer:

Imagine you slice the ring into two half arcs spanning 180 degrees each. Each has radius "R".

place the two arcs back into contact, and heat them up. for each half-hoop, the radius increases to R + delta r where delta r comes from the thermal expansion.

the diameter of the pair of arcs has been increased thereby by 2*(delta r).

From this you can see that the diameter of the hole in the ring increases when the ring is heated, thereby proving what every machinist knows: to get a firm fit of a collar on a shaft, heat the collar, slide it on over the shaft, and allow to cool. the result is called a "shrink fit".


I don't believe any of the other answers so far have mentioned something important: it is the rigid structure of the metal ring that makes this work. A gas, liquid, or a flexible solid will expand inward the way your intuition suggests.

A YouTube video of a guy blowing up a donut-shaped balloon provides an example:

Uninflated donut balloon Inflated donut balloon

  • $\begingroup$ It's not clear to me what we consider "rigid" and what we consider a "flexible solid" in this context. No solid will be perfectly rigid; and we know that this particular solid can experience thermal expansion. A donut shaped balloon isn't really a physical analog for thermal expansion; so the example doesn't really show much. $\endgroup$
    – JMac
    Apr 9, 2018 at 11:51

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